How do you find the maclaurin series expansion of #f(x) = (1 + x^2)^(1/3)#?

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Answer 1 · 2015-11-14T21:59:18

Find series for #(1+t)^(1/3)# first, then substitute #t = x^2# to find:

#f(x) = sum_(n=0)^oo (prod_(k=0)^(n-1) (1-3k))/(3^n n!)x^(2n)#

Let #g(t) = (1+t)^(1/3)#

Then:

#g^((1))(t) = 1/3(1+t)^(-2/3)#

#g^((2))(t) = -2/9(1+t)^(-5/3)#

#g^((3))(t) = 10/27(1+t)^(-8/3)#

...

#g^((n))(t) = (prod_(k=0)^(n-1) (1-3k))/3^n (1+t)^((1-3n)/3)#

and

#g^((n))(0) = (prod_(k=0)^(n-1) (1-3k))/3^n#

So:

#g(t) = g(0)/(0!) + (g^((1))(0))/(1!)t + (g^((2))(0))/(2!)t^2 + ...#

#=sum_(n=0)^oo (g^((n))(0))/(n!)t^n#

#=sum_(n=0)^oo (prod_(k=0)^(n-1) (1-3k))/(3^n n!)t^n#

Then substitute #t = x^2# to get:

#f(x) = sum_(n=0)^oo (prod_(k=0)^(n-1) (1-3k))/(3^n n!)x^(2n)#